Fight with Monsters

题目连接: Fight with Monsters

题目:

There are n monsters standing in a row numbered from 1 to n. The i-th monster has hi health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.

You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.

The fight with a monster happens in turns.

You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).

Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.

Input
The first line of the input contains four integers n,a,b and k (1≤n≤2⋅105,1≤a,b,k≤109) — the number of monsters, your attack power, the opponent’s attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h1,h2,…,hn (1≤hi≤109), where hi is the health points of the i-th monster.

Output
Print one integer — the maximum number of points you can gain if you use the secret technique optimally.

Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5

Input
1 1 100 99
100
Output
1

Input
7 4 2 1
1 3 5 4 2 7 6
Output
6

大致题意:

简单来说, 就是有A, B两个人打怪兽(A是主角), A的攻击力为a, B的攻击力为b, 如果A给予怪兽最后一击, 则加一分, 如果B给予怪兽最后一击, 则不得分. 但是因为A是主角, 他可以开挂, 效果就是可以让B不攻击一轮, 但是只能用k次. 问你最多能得多少分.

解题思路:

这个题, 有些怪物的血量刚好A打一下B打一下, 最后A打死, 不用开挂属于白给分数.
剩下的怪需要开挂能被A打死, 那就看看这个怪需要开挂多少次, 记录下来, 最后把所有开挂的次数sort一下, 取小即可.

AC代码:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int cou[1000000]; //记录打死这个怪需要开挂的次数.
int main(void)
{int n, a, b, k, h; int res = 0; //初始得分0cin >> n >> a >> b >> k;for (int i = 1; i <= n; i++) {scanf("%d", &h); //得到当前怪物血量if (h % (a + b) <= a && h % (a + b) > 0) res++; //属于白给类型, 注意判断>0的部分, 如果=0说明刚好被B打死, 得开挂else {h = h % (a + b); //打完了整数轮后怪物剩的血量, 应该是>a的if (h == 0)  h = a + b; //特殊情况, 如果B刚好打死怪物, 我们滚回一轮, 假设这轮没打cou[i] = h % a ? h / a + 1 : h / a; cou[i]--; //看看A需要打多少次打死这个怪物, 需要开挂的次数要-1,//因为最后一次A直接把怪物打死 不用开挂}}sort(cou + 1, cou + 1 + n); //排序for (int i = 1; i <= n; i++) {if (cou[i] && k - cou[i] >= 0) k -= cou[i], res++;}cout << res << endl;return 0;
}

这个题还是很水的. 没啥可说的单独记录一下, 因为最近遇到的打怪题还挺多的. 有许多的题人还有血, 怪还会回血, 有许多涉及到思路优化的地方. 我发现这些进行回合轮次战的题目, %的运用是否得当是解题的关键.

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